Here are two analogous groups, Group-I and Group-II, that list words in their decreasing order of intensity. Identify the missing word in Group-II. Group-I: Abuse -> Insult -> Ridicule. Group-II: ______ -> Praise -> Appreciate.

Correct answer: Extol. The words are arranged from stronger to weaker intensity. In the positive group, extol means to praise very highly, so it is stronger than praise and appreciate.

Had I learnt acting as a child, I ______ a famous film star. Select the most appropriate option to complete the above sentence.

Correct answer: could have been. The sentence uses inversion for an unreal past condition: 'Had I learnt...' means 'If I had learnt...'. The suitable result clause is 'could have been'.

The 12 musical notes are given as C, C#, D, D#, E, F, F#, G, G#, A, A#, and B. Frequency of each note is 12th root of 2 times the frequency of the previous note. If the frequency of the note C is 130.8 Hz, then the ratio of frequencies of notes F# and C is:

Correct answer: sqrt(2). F# is six semitone steps above C. Each step multiplies frequency by 2^(1/12), so the ratio is 2^(6/12) = sqrt(2).

The following figures show three curves generated using an iterative algorithm. The total length of the curve generated after iteration n is required. In iteration 0, the length is 1. In every iteration, each segment is replaced by 5 segments, each of length one-third of the previous segment.

Correct answer: (5/3)^n. At every iteration the number of segments is multiplied by 5 and the length of each segment is divided by 3. Hence the total length is multiplied by 5/3 each time, giving (5/3)^n.

Which one of the following plots represents f(x) = -|x|/x, where x is a non-zero real number? Note: The figures shown are representative.

Correct answer: A step plot with f(x) = 1 for x 0. For x > 0, |x|/x = 1, so f(x) = -1. For x < 0, |x|/x = -1, so f(x) = 1.

Identify the option that has the most appropriate sequence such that a coherent paragraph is formed. P: Over time, such adaptations lead to significant evolutionary changes with the potential to shape the development of new species. Q: In natural world, organisms constantly adapt to their environments in response to challenges and opportunities. R: This process of adaptation is driven by the principle of natural selection, where favorable traits increase an organism's chances of survival and reproduction. S: As environments change, organisms that can adapt their behavior, structure and physiology to such changes are more likely to survive.

Correct answer: Q -> S -> R -> P. Q introduces the general idea of adaptation. S explains survival under environmental change, R names the mechanism as natural selection, and P gives the long-term evolutionary outcome.

A stick of length one meter is broken at two locations at distances of b1 and b2 from the origin (0), as shown in the figure. Note that 0 < b1 < b2 < 1. Which one of the following is NOT a necessary condition for forming a triangle using the three pieces? Note: All lengths are in meter.

Correct answer: b1 + b2 < 1. The three pieces have lengths b1, b2 - b1, and 1 - b2. For a triangle, each piece must be less than half the total length: b1 < 0.5, b2 - b1 < 0.5, and 1 - b2 < 0.5. The condition b1 + b2 < 1 is not necessary.

Eight students (P, Q, R, S, T, U, V, and W) are playing musical chairs. The figure indicates their order or position at the start of the game. They play the game by moving forward in a circle in the clockwise direction. After the 1st round, 4th student behind P leaves the game. After 2nd round, 5th student behind Q leaves the game. After 3rd round, 3rd student behind V leaves the game. After 4th round, 4th student behind U leaves the game. Who all are left in the game after the 4th round?

Correct answer: Q; T; V; W. Following the clockwise movement and eliminating the specified students round by round leaves Q, T, V, and W after the fourth round.

The table lists the top 5 nations according to the number of gold medals won in a tournament, and also includes the number of silver and bronze medals won by them. Based only on the data provided in the table, which one of the following statements is INCORRECT? USA: Gold 40, Silver 44, Bronze 41. Canada: Gold 39, Silver 27, Bronze 24. Japan: Gold 20, Silver 12, Bronze 13. Australia: Gold 17, Silver 19, Bronze 16. France: Gold 16, Silver 26, Bronze 22.

Correct answer: USA and Canada together have less than 50% of the medals awarded to the nations in the above table.. The total medals are USA 125, Canada 90, Japan 45, Australia 52, and France 64, for a total of 376. USA and Canada together have 215 medals, which is more than 50% of 376, so that statement is incorrect.

An organization allows its employees to work independently on consultancy projects but charges an overhead on the consulting fee. The overhead is 20% of the consulting fee if the fee is up to Rs. 5,00,000. For higher fees, the overhead is Rs. 1,00,000 plus 10% of the amount by which the fee exceeds Rs. 5,00,000. The government charges a Goods and Services Tax of 18% on the total amount, that is, the consulting fee plus the overhead. An employee charges this entire amount to the client. If the client cannot pay more than Rs. 10,00,000, what is the maximum consulting fee that the employee can charge?

Correct answer: Rs. 7,24,961. For a fee F above Rs. 5,00,000, overhead = 1,00,000 + 0.1(F - 5,00,000) = 0.1F + 50,000. Total client payment is 1.18(F + overhead). Solving 1.18(1.1F + 50,000) <= 10,00,000 gives F <= about Rs. 7,24,961.

Consider the matrix A = [[2, 3, 4, 5], [0, 6, 7, 8], [0, 0, alpha, beta], [0, 0, 0, gamma]]. For which of the following combinations of alpha, beta, and gamma is the rank of A at least three? (i) alpha = 0 and beta = gamma != 0. (ii) alpha = beta = gamma = 0. (iii) beta = gamma = 0 and alpha != 0. (iv) alpha = beta = gamma != 0.

Correct answer: Only (i), (iii), and (iv). The first two rows already provide two pivots. Case (i) adds one independent row in the last column, case (iii) adds a pivot through alpha, and case (iv) gives nonzero lower diagonal entries. Case (ii) leaves rank only 2.

Consider the following series: (i) sum from n = 1 to infinity of 1/sqrt(n), (ii) sum from n = 1 to infinity of 1/[n(n+1)], and (iii) sum from n = 1 to infinity of 1/n!. Choose the correct option.

Correct answer: Only (ii) and (iii) converge. Series (i) is a p-series with p = 1/2 and diverges. Series (ii) is telescoping, and series (iii) converges to e - 1.

A pot contains two red balls and two blue balls. Two balls are drawn from this pot randomly without replacement. What is the probability that the two balls drawn have different colours?

Correct answer: 2/3. There are C(4,2) = 6 possible pairs. Favourable pairs contain one red and one blue ball, giving 2 x 2 = 4 pairs. The probability is 4/6 = 2/3.

Consider a frequency-modulated (FM) signal f(t) = Ac cos(2 pi fc t + 3 sin(2 pi f1 t) + 4 sin(6 pi f1 t)), where Ac and fc are the amplitude and frequency of the carrier waveform. The frequency f1 is in Hz, and assume that fc > 100 f1. The peak frequency deviation of the FM signal in Hz is:

Correct answer: 15 f1. The instantaneous frequency deviation is the derivative of the phase modulation term divided by 2 pi: 3 f1 cos(2 pi f1 t) + 12 f1 cos(6 pi f1 t). Its peak value is 15 f1.

Consider an additive white Gaussian noise (AWGN) channel with bandwidth W and noise power spectral density N0/2. Let Pav denote the average transmit power constraint. Which one of the following plots illustrates the dependence of channel capacity C on bandwidth W, keeping Pav and N0 fixed?

Correct answer: Capacity increases with bandwidth and approaches a finite limiting value.. For fixed average power, C = W log2(1 + Pav/(N0 W)). As W increases, capacity increases monotonically but approaches the finite limit Pav/(N0 ln 2).

The Nyquist plot of a system is given. Let omegaP, omegaQ, omegaR, and omegaS be the positive frequencies at the points P, Q, R, and S, respectively. Which one of the following statements is TRUE?

Correct answer: omegaS is the gain crossover frequency and omegaQ is the phase crossover frequency. The gain crossover frequency occurs where the Nyquist plot intersects the unit circle. The phase crossover frequency occurs where it intersects the negative real axis. From the given plot these are S and Q, respectively.

Consider the discrete-time system with input x[n] and output y[n]. In the figure, h1[n] and h2[n] denote the impulse responses of LTI Subsystems 1 and 2, respectively. Also, delta[n] is the unit impulse, and b > 0. Assuming h2[n] is not equal to delta[n], the overall system is:

Correct answer: nonlinear and time variant. The added b delta[n] and -b delta[n] terms create an additive response component independent of the input. Since h2[n] is not delta[n], this component does not cancel, so the system is not linear and is also time variant.

Consider a continuous-time, real-valued signal f(t) whose Fourier transform F(omega) = integral from -infinity to infinity of f(t) exp(-j omega t) dt exists. Which one of the following statements is always TRUE?

Consider a part of an electrical network as shown. Some node voltages and the current flowing through the 3 ohm resistor are indicated. The voltage, in volts, at node X is:

Correct answer: 20/3. The indicated current through the 3 ohm resistor gives the adjacent node voltage as 9 - 3 = 6 V. The branch from 8 V to 6 V has total resistance 2 + 1 = 3 ohm, so the current is 2/3 A. Hence VX = 8 - 2(2/3) = 20/3 V.

Let iC, iL, and iR be the currents flowing through the capacitor, inductor, and resistor, respectively, in the circuit given below. The AC admittances are given in Siemens. Which one of the following is TRUE?

Correct answer: iC = 0.25 angle 180 degrees A, iL = 0.1 angle 0 degrees A, iR = 0.2 angle 90 degrees A. With source voltage 1 angle 90 degrees V, branch currents are YV. Thus j0.25 times the source gives iC = 0.25 angle 180 degrees A, -j0.1 gives iL = 0.1 angle 0 degrees A, and 0.2 gives iR = 0.2 angle 90 degrees A.

A simplified small-signal equivalent circuit of a BJT-based amplifier is given. The small-signal voltage gain vo/vs in V/V is:

Correct answer: -beta RL / (RS + rpi). The input current is ib = vs/(RS + rpi). The controlled collector current is beta ib, and the output across RL is inverted, so vo/vs = -beta RL/(RS + rpi).

The ideal BJT in the circuit is biased in the active region with beta = 100. If IB is 10 microampere, then VCE in volts, rounded off to two decimal places, is:

Correct answer: 1.92. IC = beta IB = 1 mA and IE = 1.01 mA, so VE = 1.01 mA x 3 kohm = 3.03 V. The collector resistor carries IC + IB = 1.01 mA, so VC = 10 - 1.01 mA x 5 kohm = 4.95 V. Hence VCE = 4.95 - 3.03 = 1.92 V.

A 3-input majority logic gate has inputs X, Y, and Z. The output F of the gate is logic 1 if two or more of the inputs are logic 1. The output F is logic 0 if two or more of the inputs are logic 0. Which one of the following options is a Boolean expression of the output F?

Correct answer: XY + YZ + ZX. A 3-input majority output is 1 for minterms 3, 5, 6, and 7. The minimized expression is XY + YZ + ZX.

A full adder and an XOR gate are used to design a digital circuit with inputs X, Y, and Z, and output F. The input Z is connected to the carry-in input of the full adder. If input Z is set to logic 1, then the circuit functions as ________ with X and Y as inputs.

Correct answer: a subtractor. With Z = 1, the XOR gate complements Y and the full adder receives carry-in 1. This implements X + Y' + 1, which is the two's-complement subtraction X - Y.

Consider the function f: R -> R defined as f(x) = 2x^3 - 3x^2 - 12x + 1. Which of the following statements is/are correct? Here, R is the set of real numbers.

Correct answer: A and B. The cubic has positive leading coefficient, so it is unbounded above and below; hence no global maximizer or minimizer exists. Also f' = 6(x - 2)(x + 1), so x = -1 is a local maximum and x = 2 is a local minimum.

Consider the unity-negative-feedback system shown, where gain K >= 0. The root locus of this system is shown in the figure. For what value(s) of K will the system have a pole at -1 + j1?

Correct answer: For no positive value of K. The point -1 + j1 is not on the positive-gain root locus shown. Therefore no positive value of K places a closed-loop pole at that point.

Let x[n] be a discrete-time signal whose z-transform is X(z). Which of the following statements is/are TRUE?

Correct answer: C and D. A DTFT need not always exist. The ROC cannot contain poles but may contain zeros, so statement B is false. The DTFT exists when the unit circle is in the ROC, and alpha delta[n] has a constant z-transform with the entire z-plane as ROC.

Consider a message signal m(t) bandlimited to [-W, W]. For DSB-SC, fDSB(t) = Ac m(t) cos(2 pi fc t). For AM, fAM(t) = Ac(1 + mu m(t)) cos(2 pi fc t). Consider the statements: (i) An envelope detector can be used for demodulation in the DSB-SC scheme if m(t) > 0 for all t. (ii) An envelope detector can be used for demodulation in the AM scheme only if m(t) > 0 for all t. Which of the following options is/are correct?

Correct answer: A and D. For DSB-SC, if m(t) is always positive, the envelope is proportional to m(t), so statement (i) is true. For AM, envelope detection requires 1 + mu m(t) to remain nonnegative, not m(t) > 0 for all t; hence statement (ii) is false.

Which of the following statements is/are TRUE with respect to an ideal opamp?

Correct answer: A and C. An ideal op-amp has infinite input resistance and infinite open-loop differential gain. Its output resistance is zero and its common-mode gain is zero.

Which of the following statements is/are TRUE with respect to ideal MOSFET-based DC-coupled single-stage amplifiers having finite load resistors?

Correct answer: B and D. An ideal MOSFET gate draws no current, so the common-source input resistance is infinite. The common-source amplifier is inverting, while the common-drain source follower is non-inverting. A common-gate amplifier has a finite small-signal input resistance at the source.

Which of the following can be used as an n-type dopant for silicon? Select the correct option(s).

Correct answer: A and D. n-type silicon is formed using pentavalent donor impurities. Arsenic and phosphorous are group-V dopants, while boron and gallium are trivalent p-type dopants.

The function y(t) satisfies t^2 y''(t) - 2t y'(t) + 2y(t) = 0, where y'(t) and y''(t) denote the first and second derivatives of y(t), respectively. Given y'(0) = 1 and y'(1) = -1, the maximum value of y(t) over [0, 1] is ________ rounded off to two decimal places.

Correct answer: 0.25. The Cauchy-Euler equation has solution y = C1 t + C2 t^2. From y'(0) = 1, C1 = 1. From y'(1) = -1, C1 + 2C2 = -1, so C2 = -1. Hence y = t - t^2, whose maximum on [0,1] is 0.25.

The generator matrix of a (6,3) binary linear block code is G = [[1,0,0,1,0,1], [0,1,0,0,1,1], [0,0,1,1,1,0]]. The minimum Hamming distance dmin between codewords equals ________.

Correct answer: 3. For a linear block code, dmin is the minimum Hamming weight among all nonzero codewords. The nonzero combinations of the generator rows have minimum weight 3.

All the components in the bandpass filter are ideal. The lower -3 dB frequency of the filter is 1 MHz. The upper -3 dB frequency in MHz, rounded off to the nearest integer, is ________.

Correct answer: 50. The circuit can be separated into first-order high-pass and low-pass factors. Using the given lower -3 dB frequency of 1 MHz and the component ratios in the ideal filter gives the upper -3 dB frequency as 50 MHz.

A 4-bit weighted-resistor DAC with inputs b3, b2, b1, and b0 (MSB to LSB) is designed using an ideal opamp. The switches are closed when the corresponding input bits are logic 1 and open otherwise. When the input b3b2b1b0 changes from 1110 to 1101, the magnitude of the change in the output voltage Vo in mV, rounded off to the nearest integer, is ________.

Correct answer: 250. GATE Shortcut Observe the bit change: 1110 -> 1101 Only two bits change: b1 : 1 -> 0 b0 : 0 -> 1 Change due to b1: Delta Vb1 = 2 x 1/4 = 0.5 V Change due to b0: Delta Vb0 = 2 x 1/8 = 0.25 V Net change: 0.5 - 0.25 = 0.25 V 250 mV Key Formula Vo = -VREF (b3 + b2/2 + b1/4 + b0/8) Answer: 250 mV

Let G(s) = 1/(s^2 - 1) be the transfer function of a second-order system. A controller M(s) is connected to the system G(s) in the unity negative feedback loop. Consider the statements: (i) There exists no controller of the form M(s) = KI/s, where KI is a positive real number, such that the closed-loop system is stable. (ii) There exists at least one controller of the form M(s) = KP + sKD, where KP and KD are positive real numbers, such that the closed-loop system is stable. Which option is correct?

Correct answer: (i) is FALSE and (ii) is TRUE. With M(s)=KI/s, the characteristic equation is s^3 - s + KI = 0, which cannot be made Hurwitz stable for KI > 0. With M(s)=KP+sKD, the characteristic equation is s^2 + KD s + KP - 1 = 0, which is stable for KD > 0 and KP > 1.

Consider the polynomial p(s) = s^5 + 7s^4 + 3s^3 - 33s^2 + 2s - 40. Let L, I, R be defined as follows: L is the number of roots of p(s) with negative real parts; I is the number of roots of p(s) that are purely imaginary; R is the number of roots of p(s) with positive real parts. Which option is correct?

Correct answer: L = 2, I = 2, and R = 1. The polynomial factors as (s - 2)(s + 4)(s + 5)(s^2 + 1). Thus it has two roots with negative real parts, two purely imaginary roots, and one root with positive real part.

Consider a continuous-time finite-energy signal f(t) whose Fourier transform vanishes outside [-omega_c, omega_c]. The signal f(t) is uniformly sampled to obtain y(t) = f(t)p(t), where p(t) = sum from n = -infinity to infinity of delta(t - tau - nTs), with Ts > 0 and tau > 0. The sampled signal y(t) is passed through an ideal lowpass filter h(t) = omega_c Ts sin(omega_c t)/(pi omega_c t), with cutoff frequency omega_c and passband gain Ts. The output of the filter is:

Correct answer: f(t - tau) if Ts < pi/omega_c. The shifted sampling train introduces a phase factor exp(-j omega tau) in the baseband spectrum. With no aliasing, requiring Ts < pi/omega_c, the reconstruction filter output is f(t - tau).

In the circuit, M1 is an ideal AC voltmeter and M2 is an ideal AC ammeter. The source voltage is vs(t) = 100 cos(200t). What should be the value of the variable capacitor C such that the RMS readings on M1 and M2 are 25 V and 5 A, respectively?

Correct answer: 25 microfarad. The meter readings require the load impedance magnitude to be 25/5 = 5 ohm. Since the series resistance is already 5 ohm, the net reactance must be zero. Hence omega L = 1/(omega C), so C = 1/(200^2 x 1) = 25 microfarad.

The Z-parameter matrix of a two port network relates the port voltages and port currents as follows: [V₁] [V₂] = Z [I₁] [I₂] The Z-parameter matrix (with each entry in Ohms) of the network shown below is ________.

Correct answer: [[10/3, 2/3], [2/3, 10/3]]. Quick GATE Shortcut (Using Y-Parameters) For this ladder network: 1. Write nodal admittance matrix: Y = [ 0.3125 -0.0625 ] [ -0.0625 0.3125 ] 2. Invert the matrix: Z = Y⁻¹ 3. Result: Z = [ 10/3 2/3 ] Ω [ 2/3 10/3 ] Key Formula [ V₁ ] [ Z₁₁ Z₁₂ ] [ I₁ ] [ V₂ ] = [ Z₂₁ Z₂₂ ] [ I₂ ] ✅ Final Z-parameter matrix [ 3.33 0.67 ] Ω [ 0.67 3.33 ]

A source transmits symbol S that takes values uniformly at random from the set {-2, 0, 2}. The receiver obtains Y = S + N, where N is a zero-mean Gaussian random variable independent of S. The receiver uses the maximum likelihood decoder to estimate the transmitted symbol S. Suppose the probability of symbol estimation error Pe is expressed as Pe = alpha P(N > 1), where P(N > 1) denotes the probability that N exceeds 1. What is the value of alpha?

Correct answer: 4/3. The ML thresholds are at -1 and 1. Error probabilities for S = -2, 0, and 2 are P(N > 1), 2P(N > 1), and P(N > 1), respectively. Averaging over three equally likely symbols gives Pe = (4/3)P(N > 1).

Consider a real-valued random process f(t) = sum from n = 1 to N of an p(t - nT), where T > 0 and N is a positive integer. Here, p(t) = 1 for t in [0, 0.5T] and 0 otherwise. The coefficients an are pairwise independent, zero-mean, unit-variance random variables. Read the statements and choose the correct option. (i) The mean of f(t) is independent of time t. (ii) The autocorrelation function E[f(t)f(t + tau)] is independent of time t for all tau.

Correct answer: (i) is TRUE and (ii) is FALSE. Since every coefficient has zero mean, E[f(t)] = 0 for all t. The autocorrelation depends on whether t and t + tau fall inside the rectangular pulses, so it is not independent of t.

The identical MOSFETs M1 and M2 in the circuit are ideal and biased in the saturation region. M1 and M2 have transconductance gm of 5 mS. The input signals are V1 = 2.5 + 0.01 sin(omega t) and V2 = 2.5 - 0.01 sin(omega t). The output signal V3 in volts is:

Correct answer: 4 - 0.05 sin(omega t). GATE Shortcut For a differential pair: vₒ = - (gₘRᴅ / 2) vᵢd Here, gₘRᴅ = (5 mS)(1 kΩ) = 5 and vᵢd = 0.02 sinωt Thus, vₒ = - (5 / 2)(0.02 sinωt) = -0.05 sinωt Adding the DC drain voltage: V₃ = 4 - 0.05 sinωt Key Formula vₒ = - (gₘRᴅ / 2) vᵢd ✅ Answer: (D) 4 - 0.05 sinωt

A 10-bit analog-to-digital converter (ADC) has a sampling frequency of 1 MHz and a full-scale voltage of 3.3 V. For an input sinusoidal signal with frequency 500 kHz, the maximum SNR in dB, rounded off to two decimal places, and the data rate in Mbps at the output of the ADC are, respectively:

Correct answer: 61.96 and 10. For an ideal N-bit ADC, maximum SQNR for a full-scale sinusoid is 6.02N + 1.76 dB. With N = 10, this is 61.96 dB. The output data rate is 10 bits/sample x 1 Msamples/s = 10 Mbps.

A positive-edge-triggered sequential circuit is shown. There are no timing violations. Input P0 is set to logic 0 and P1 is set to logic 1 at all times. The timing diagram of inputs SEL and S is also shown. The sequence of output Y from time T0 to T3 is:

Correct answer: 1011. Tracing the two positive-edge-triggered D flip-flops at T0, T1, T2, and T3 using the shown SEL and S waveforms gives the output sequence 1011.

The intrinsic carrier concentration of a semiconductor is 2.5 x 10^16 per cubic meter at 300 K. If the electron and hole mobilities are 0.15 m^2/Vs and 0.05 m^2/Vs, respectively, then the intrinsic resistivity of the semiconductor in kilo-ohm meter at 300 K is:

Correct answer: 1.25. Intrinsic conductivity is sigma = q ni(mu_n + mu_p) = 1.6e-19 x 2.5e16 x 0.20 = 8e-4 S/m. Thus resistivity is 1/sigma = 1250 ohm-m = 1.25 kilo-ohm-m.

In the circuit shown, identical transistors Q1 and Q2 are biased in the active region with beta = 120. The Zener diode is in the breakdown region with VZ = 5 V and IZ = 25 mA. If IL = 12 mA and VEB1 = VEB2 = 0.7 V, then the values of R1 and R2 in kilo-ohm, rounded off to one decimal place, are respectively:

Correct answer: 0.6 and 0.4. Using the active-region BJT current relations with beta = 120 and the specified Zener current fixes the branch currents. The corresponding resistor drops give R1 = 0.6 kilo-ohm and R2 = 0.4 kilo-ohm.

The electron mobility mu_n in a non-degenerate germanium semiconductor at 300 K is 0.38 m^2/Vs. The electron diffusivity D_n at 300 K in cm^2/s, rounded off to the nearest integer, is:

Correct answer: 98. Using Einstein's relation D_n = mu_n kT/q gives D_n = 0.38 x (1.38e-23 x 300)/(1.6e-19) = 0.00983 m^2/s = 98.3 cm^2/s.

A square metal sheet of 4 m x 4 m is placed on the x-y plane as shown. If the surface charge density on the sheet is rho_s(x,y) = 4|y| microcoulomb per square meter, then the total charge in microcoulomb, rounded off to the nearest integer, on the sheet is:

Correct answer: 64. GATE Shortcut The charge density depends only on y, so: 1. Average value of 4|y| over -2 <= y <= 2: Average |y| = (1/4) ∫_{-2}^{2} |y| dy = 4/4 = 1 Thus average charge density: ρ_avg = 4 × 1 = 4 μC/m^2 2. Area of sheet: A = 4 × 4 = 16 m^2 3. Total charge: Q = ρ_avg A Q = 4 × 16 Q = 64 μC Key Formula Q = ∬_S ρ_s dS Topic: Electrostatics – Surface Charge Distribution Answer: 64 μC ✅

An electric field of 0.01 V/m is applied along the length of a copper wire of circular cross-section with diameter 1 mm. Copper has conductivity 5.8 x 10^7 S/m. The current in amperes, rounded off to two decimal places, flowing through the wire is:

Correct answer: 0.46. Current density J = sigma E = 5.8e7 x 0.01 = 5.8e5 A/m^2. The cross-sectional area is pi(0.5e-3)^2, so the current is J A = 0.455 A, which rounds to 0.46 A.

Consider a non-negative function f(x), continuous and bounded over [2,8]. Let M and m denote, respectively, the maximum and minimum values of f(x) over the interval. Among the combinations of alpha and beta below, choose the one(s) for which beta <= integral from 2 to 8 of f(x) dx <= alpha is guaranteed to hold.

Correct answer: beta = 5m, alpha = 7M. Since f is non-negative and bounded between m and M on an interval of length 6, 6m <= integral <= 6M. Therefore beta = 5m is safely below the integral and alpha = 7M is safely above it.

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